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Question Number 214675 by cherokeesay last updated on 16/Dec/24

Answered by mr W last updated on 16/Dec/24

Commented by mr W last updated on 16/Dec/24

$$x^2+y^2=2^2=4$$$$AB=\sqrt{x^2+9y^2}=\sqrt{x^2+9(4-x^2)}=\sqrt{36-8x^2}$$$$CD=\sqrt{y^2+9x^2}=\sqrt{4-x^2+9x^2}=\sqrt{4+8x^2}$$$$\frac{p}{AB}=\frac{q}{CD}=\frac{2}{6-2}=\frac{1}{2}$$$$p=\frac{\sqrt{36-8x^2}}{2}=\sqrt{9-2x^2}$$$$q=\frac{\sqrt{4+8x^2}}{2}=\sqrt{1+2x^2}$$$$p^2+q^2-2pq\cos 45°=2^2$$$$9-2x^2+1+2x^2-\sqrt{2(9-2x^2)(1+2x^2)}=4$$$$\sqrt{2(9-2x^2)(1+2x^2)}=6$$$$4x^4-16x^2+9=0$$$$x^2=2+\frac{\sqrt{7}}{2}$$$$y^2=2-\frac{\sqrt{7}}{2}$$$$xy=\sqrt{(2+\frac{\sqrt{7}}{2})(2-\frac{\sqrt{7}}{2})}=\frac{3}{2}$$$$\left[ABCD\right]=\frac{4x\times 4y}{2}=8xy=8\times \frac{3}{2}=12$$

Commented by cherokeesay last updated on 16/Dec/24

$$great!thankyoumaster!$$

Commented by som(math1967) last updated on 17/Dec/24

$$Siranglebetween3xand3yis$$$$90°someetingpointofthemis$$$$circumcentreoftriangle$$$$so3x=3y?$$

Commented by mr W last updated on 17/Dec/24

$$nosir.itmeansonlythatthemeeting$$$$pointisonthecircumcirclewith$$$$ADasdiameter.$$

Commented by som(math1967) last updated on 17/Dec/24

$$Oksirthankyou$$

Commented by Matica last updated on 18/Dec/24

$$Excuseme.ididntunderstandhowyougot6-2.pleasehelpexplainthat.thanks.$$

Commented by mr W last updated on 18/Dec/24

Commented by mr W last updated on 18/Dec/24

$$\frac{p}{A^{\prime }C}=\frac{p}{AB}=\frac{q}{CD}=\frac{BC}{A^{\prime }D}=\frac{2}{6-2}$$

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